A convergence criterion for sums based on an alternating sequence.

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For some particular cases, we nevertheless are able to show
convergence.

**Theorem 1** (Leibniz Convergence Criterion). *Let
\((a_n)_{n\in\mathbb{N}}\) be
a monotonically decreasing real sequence with \[\lim_{n\to\infty}a_n=0.\] Then the
alternating sequence \[\sum_{k=1}^\infty
(-1)^ka_k\] converges.*

*Proof:* Since \((a_n)_{n\in\mathbb{N}}\) is convergent to
zero and monotonically decreasing, we have \(a_n\geq0\) for all \(n\in\mathbb{N}\). Let \((s_n)_{n\in\mathbb{N}}\) be the
corresponding sequence of partial sums. Then we have for all \(l\in \mathbb{N}\) that \[s_{2l+2}-s_{2l}=-a_{2l+1}+a_{2l+2}\leq 0,\quad
s_{2l+3}-s_{2l+1}=a_{2l+2}-a_{2l+3}\geq 0,\] i.e., the
subsequence \((s_{2l})_{l\in\mathbb{N}}\) is
monotonically decreasing and \((s_{2l+1})_{l\in\mathbb{N}}\) is
monotonically increasing. Furthermore, due to \(s_{2l+1}-s_{2l}=-a_{2l+1}\leq0\) holds
\[s_{2l+1}\leq s_{2l}.\]

Altogether, the \((s_{2l})_{l\in\mathbb{N}}\) is
monotonically decreasing and bounded from below, and \((s_{2l+1})_{l\in\mathbb{N}}\) is
monotonically increasing and bounded from above. By the Theorem on
monotonic and bounded sequences, both subsequences are convergent. Due
to \[\lim_{l\to\infty}(s_{2l+1}-s_{2l})=\lim_{l\to\infty}a_{2l+1}=0,\]
an application of the formulae for convergent sequences yields that both
subsequence have the same limit, i.e., \[\lim_{l\to\infty}s_{2l+1}=\lim_{l\to\infty}s_{2l}=s\]
for some \(s\in\mathbb{R}\). Let \(\varepsilon>0\). Then there exists some
\(N_1\) such that for all \(l\geq N_1\) holds \(|s_{2l}-s|<\varepsilon\). Furthermore,
there exists some \(N_2\) such that for
all \(l\geq N_2\) holds \(|s_{2l+1}-s|<\varepsilon\). Now choosing
\(N=\max\{2N_1,2N_2+1\}\), we can say
the following for some \(m\geq
N\):

In the case where \(m\) is even, we
have some \(l\in\mathbb{N}\) with \(m=2l\). By the choice of \(N\), we also have \(l\geq N_1\) and thus \[|s_{m}-s|=|s_{2l}-s|<\varepsilon.\] In
the case where \(m\) is odd, we have
some \(l\in\mathbb{N}\) with \(m=2l+1\). By the choice of \(N\), we also have \(l\geq N_2\) and thus \[|s_{m}-s|=|s_{2l+1}-s|<\varepsilon.\]
\(\Box\)

**Example 2**.

The *alternating harmonic series* \[\sum_{k=1}^\infty \frac{(-1)^k}{k}.\]
converges. The limit is (without proof) \(\log(2)\).

The *alternating Leibniz series* \[\sum_{k=1}^\infty \frac{(-1)^k}{2k+1}.\]
converges. The limit is (without proof) \(\frac\pi4\).

The series \[\sum_{k=1}^\infty
\frac{(-1)^k}{\sqrt{k}}.\] converges. The limit is not
expressible in a closed form.

We will later treat the topic of *Taylor series*.
Thereafter we will be able to determine some further limits of
sequences.

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